Question

How Many terms of the A.P:18,16,14......be taken so that their sum is zero

Answer 1

AP : 18, 16, 14, ......

First term (a) = 18

Common difference (d) = 16 - 18 = - 2

Sum of the terms (Sn) = 0

Number of terms for which sum is zero = (n) = ?

Now,

Sn = n/2 × [2a + (n - 1)d]

=> 0 = n/2 × [2×18 + (n - 1)×(- 2)]

=> 0 = n(36 - 2n + 2)

=> n(38 - 2n) = 0

=> 38 - 2n = 0

=> 38 = 2n

=> n = 38/2 = 19

Therefore, sum of 19 terms of the AP will be zero.

Answer 2

Answer:

$Consider \: the \: \: given \: A.P. \: series.</p><p> \\ </p><p>27,24,21,......</p><p></p><p> \\ </p><p>Here, a=27,d=−3</p><p></p><p> \\ </p><p>Since, Sum=0</p><p> \\ </p><p></p><p>Therefore,</p><p></p><p>$

$sum = \frac{n}{2} [2a + (n - 1)d]$

$0= \frac{n}{2} [2 \times 27 + (n - 1) \times - 3]$

$54 - 3n + 3 = 0$

$57 - 3n = 0$

$57 = 3n$

$n = \frac{ \cancel{ 57}}{ \cancel 3} = 19$

$so, \: \boxed{n = 19}$